Course Introduction · Course Introduction • Instructor Dr. Cheng-Yu Lai SCS300 • CHEM 102- 3...
Transcript of Course Introduction · Course Introduction • Instructor Dr. Cheng-Yu Lai SCS300 • CHEM 102- 3...
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Course Introduction• Instructor Dr. Cheng-Yu Lai SCS300• CHEM 102- 3 Credits Lecture and/or 1 credit Lab.• Textbook• Evaluation and Assessment
100-90=A (GPA=4.0)89-80=B (GPA=3.0)79-70=C (GPA=2.0)69-60=D (GPA=1.0)59-0=F (GPA=0.0)
Attendance and performance + Quizzes /Homework (30%) ;3 Exams (50%) + 1ACS Test (20%)= 100%
Any Question ?
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Chapter 11 Liquids, Solids andIntermolecular Forces
• Phase Diagrams• Intermolecular Forces• Types of Substances
- Molecular, Covalent,ionic and metallic Solids
• Crystal Structures• Vapor Pressure and Phase
Equilibrium
Modified by Dr. Cheng-Yu Lai
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3 Phases of water-Phase Transition
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What is Phase Diagram ?
• Phase diagrams are graphicalrepresentations of the pressure andtemperature dependence of a puresubstance– Pressure on the y-axis– Temperature on the x-axis
• Three places to consider– In a region, one phase exists– On a line, two phases exist in
equilibrium– At a point, three phases exist in
equilibrium
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Phase Diagram of Water• Curve b (green) is the vapor pressure-
temperature curve of liquid water• Curve c (red) is the vapor pressure curve
of ice• Line a (blue) gives the temperature-
pressure dependence for ice inequilibrium with water
• Point X is the triple point– All three phases are in equilibrium– There is only one triple point for a
pure substance– For water, the triple point is at 0.01
°C and 4.56 mmHg
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Definition of Phase ConversionSublimation of I2
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Phase Changes
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Chapter 10/8© 2012 Pearson Education, Inc.
Phase Changes
Heat (Enthalpy) of Fusion (DHfusion ): The amountof energy required to overcome enoughintermolecular forces to convert a solid to a liquid
Heat (Enthalpy) of Vaporization (DHvap): Theamount of energy required to overcome enoughintermolecular forces to convert a liquid to a gas
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Heating Curve of Water
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© 2014 Pearson Education, Inc.
Segment 1
• Heating 1.00 mole of ice at −25.0°C up to themelting point, 0.0°C
• q = mass× Cs× ∆T– Mass of 1.00 mole of ice = 18.0 g– Cs = 2.09 J/mol ∙°C
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© 2014 Pearson Education, Inc.
Segment 2
• Melting 1.00 mole of ice at the melting point,0.0°C
• q = n ∙ ∆Hfus– n = 1.00 mole of ice– ∆Hfus = 6.02 kJ/mol
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© 2014 Pearson Education, Inc.
Segment 3
• Heating 1.00 mole of water at 0.0 °C up tothe boiling point, 100.0°C
• q = mass× Cs× ∆T– Mass of 1.00 mole of water = 18.0 g– Cs = 2.09 J/mol ∙°C
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© 2014 Pearson Education, Inc.
Segment 4
• Boiling 1.00 mole of water at the boilingpoint, 100.0°C
• q = n ∙ ∆Hvap– n = 1.00 mole of ice– ∆Hfus = 40.7 kJ/mol
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© 2014 Pearson Education, Inc.
Segment 5
• Heating 1.00 mole of steam at 100.0 °C upto 125.0°C
• q = mass× Cs× ∆T– Mass of 1.00 mole of water = 18.0 g– Cs = 2.01 J/mol ∙°C
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Critical Temp and Pressure
For every liquid, there is a temperature abovewhich only vapor can exist. This is the criticaltemperature. At this temperature, thepressure is called the critical pressure.Together, the critical temperature and pressureare called the critical point
Tubes containing water atseveral temperatures.Note that at or above374oC (the criticaltemperature for water),only water vapor exists inthe tube.
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The slope of the solid-liquid line depicts the behavior of the freezingpoint as pressure is increased or decreased
Positive slope: solid is denser than liquidNegative slope: liquid is denser than solid – water due to H bonding
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Intermolecular Forces
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Intermolecular Forces – Among MoleculesOne could explain the observed phase-change behavior ofdifferent materials. For example, water (18 g/mol) is a liquidat ambient conditions but methane (16 g/mol) is a gas. Theexplanation requires an understanding of the intermolecularforces that are involved
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2 Categories/3 Types Intermolecular Forces
London Force, Dipole-Dipole Interaction and Hydrogen Bond
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1. London Dispersion Forces
Fritz London1900-1954
London forces increase with thesize of the molecules.
Synonyms: “London forces”, “dispersion forces”, and“dispersion-interaction forces”
1. All substances have dispersion forces.Also called London or Van Der Waals forcesStem from induced dipoles in molecules
2. Motion of electrons in the moleculecauses transient dipoles to formAs molar mass increases, dispersion forcesbecome stronger,
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London Forces in Hydrocarbons
MW↑ BP ↑
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2. Dipole-Dipole Interaction
1. Molecules with permanent dipoles display dipole forcesDispersion forces are also present but are much weaker2. Adjacent molecules line up so that the negative pole of onemolecule is as close as possible to the positive pole of anothermolecule A molecule has dipole–dipole forces if it is polar.
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© 2014 Pearson Education, Inc.Chemistry: A Molecular Approach, 3rd EditionNivaldo J. Tro
Example 11.1 Dipole–Dipole Forces
A molecule has dipole–dipole forces if it is polar. To determine if a molecule is polar, (1) determine if themolecule contains polar bonds and (2) determine if the polar bonds add together to form a net dipole moment(Section 9.6).
(a) CO2
(1) Since the electronegativity of carbonis 2.5 and that of oxygen is 3.5(Figure 9.8), CO2 has polar bonds.
(2) The geometry of CO2 is linear.Consequently, the dipoles of the polarbonds cancel, so the molecule is notpolar and does not havedipole–dipole forces.
Solution
Which of these molecules have dipole–dipole forces?a. CO2 b. CH2CI2 c. CH4
FIGURE 9.8 Electronegativities of the Elements Electronegativitygenerally increases as we move across a row in the periodic table anddecreases as we move down a column.
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3. Hydrogen BondingUnusually strong type of dipole force H attached to a(n) N,O, or F
HF, H2O and NH3: unusually high boiling points as a result ofhydrogen bonding
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Chapter 10/25 © 2012 Pearson Education, Inc.
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Boiling point as a measure of intermolecularattractive forces
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Chapter 10/28© 2012 Pearson Education, Inc.
Intermolecular Forces
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Chapter 10/29© 2012 Pearson Education, Inc.
Kinds of Solids
Amorphous Solids: Particles are randomly arranged andhave no ordered long-range structure. Example - rubber
Crystalline Solids: Particles have an orderedarrangement extending over a long range.• Ionic solids• Molecular solids• Covalent network solids• Metallic solids
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4 Types of Solid Interaction
SiO2
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Kinds of Solids
1. Ionic Solids: Particles are ions ordered in aregular three-dimensional arrangement and heldtogether by ionic bonds. Example - sodium chloride.
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Ionic Solids-Oppositely-charged ions held togetherby strong electrical forces
Strengths of Ionic BondsCoulomb’s Law
Strength of ionic bond depends on1. Charges of the ions (higher charges produce strongerbonds); Q2. Sizes of the ions (smaller internuclear distances resultin stronger bonds) ; d
anioncation rrdd
QQkE
21
• Characteristics– Nonvolatile; high melting points (600-2000 °C)– Nonconductors of electricity in the solid state
• Conduct when melted or dissolved in water– Many are soluble in water but not in nonpolar solvents
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Strength of ionic bond
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Chapter 10/34© 2012 Pearson Education, Inc.
Kinds of Solids
2. Molecular Solids: Particles are molecules heldtogether by intermolecular forces. Example - ice
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3. Network Covalent Solids• Characteristics
– High melting points, often above 1000 °C– Covalent bonds must be broken to melt the substance
• Examples– Graphite and diamond: allotropes
• Diamond is three-dimensional and tetrahedral• Graphite is two-dimensional and planar
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4. Metallic Solids - Structural unit are +1, +2 and +3metals with associated electrons
• Characteristics of metals– High electrical conductivity
• Highly mobile electrons in structure– High thermal conductivity
• Heat is carried through the structure by collision betweenelectrons
– Ductility and malleability• Can be drawn into wire or hammered into sheets
– Luster• Polished metal surfaces reflect light
– Insolubility in water and other common solvents
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Example
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Crystal Building Blocks- atoms/ ions arrangements
• Crystals have definite geometric formsbecause the atoms or ions are arranged indefinite, three-dimensional patterns
• Metals crystallize into one of three unit cells1. Simple cubic (SC): eight atoms at the corners2. Face centered cubic (FCC): simple cubic plus one
atom in the center of each face3. Body-centered cubic (BCC): simple cubic plus
one atom in the center of the cube
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Crystal Structures – Cubic Unit Cells
Simple Face-Centered Body-Centered
a
aa
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Simple Cubic
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Body-Centered Cubic
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Face-Centered Cubic
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© 2014 Pearson Education, Inc.
Cesium Chloride Structures
• Coordination number = 8• ⅛ of each Cl─ (184 pm)
inside the unit cell• Whole Cs+ (167 pm) inside
the unit cell– Cubic hole = hole in simple
cubic arrangement of Cl─ ions• Cs:Cl = 1: (8× ⅛); therefore
the formula is CsCl.
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Atom Numbers Per Unit- Empirical Formula
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Side of cell (s) and radius of atom or ion (r)
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Property of Cubic Unit Cells• Three other ways to look at the crystalline unit cells:
1. Number of atoms per unit cell• SC: 1 FCC: 4 BCC: 2
2. Relation between side of cell (s) and radius of atom or ion (r)• SC: FCC: BCC:
3. Percentage of empty space• SC: 47.5 FCC: 32.0 BCC: 26.0
sr 2 24 sr 34 sr
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Chapter 10/47© 2012 Pearson Education, Inc.
Vapor Pressure
Vapor Pressure: The partial pressure of a gas inequilibrium with liquid at a constant temperature
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Vapor Pressure and Phase Equilibrium
Molecules are entering the vapor phase from the liquidand condensing from the vapor phase to the liquid at thesame rate
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Vapor Pressure and Volume / Temperature
The pressure of vaporin equilibrium with aliquid is called vaporpressure.
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Vapor Pressure Versus TemperatureThe vapor pressure of a liquid increases asthe temperature rises
Increase in P is not linear withtemperatureWater
VP is 24 mmHg at 25 °CVP is 92 mmHg at 50 °C
To make a linear plot, the natural logarithmis required
Clausius-Clapeyron Equation
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The Clausius-Clapeyron Equation
bRTH
P vap ln
• For many purposes, a two-point equation is useful– Two pressures– Two temperatures
• Notes:– Temperatures must be in Kelvin
• R = 8.31 J/mol·K• ΔHvap must be in J for use with this value for R
211
2 11lnTTR
HPP vap
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The Clausius-Clapeyron Equation
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Example
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Example 9.8, (Cont’d)
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Example 9.8, (Cont’d)